If the number 5m15n, where m and n represent the thousands’ and unit

Since the number has to be divisible by 36 => 2^2 and 3^2
so lets check how it can be made divisible by 9 first
5+m+1+n+5 = 11+m+n
For it to be divisble by 9 => m+n = 7(ie. 18-11) or 16(ie. 27-11)
m-n has to be maximised so if we consider 16 then m & n (in any order) can be {8,8}, {9,7} the latter case is not possible since the number has to be divisible by 4 also and in units's place we cannot have odd number.
Also 0 is not maximum value (in case m,n are 8,8) so lets ignore it.
Lets see how we can get 7 from m ,n => {3,4},{6,1} {2,5} in any order
since 6-1 =5 is maximum value , we can check it for divisibility
lets check 51156 it is clearly divisible by 4 from last two digits. Hence |m-n| max = 5
Option C.

Experts please let me know if the above solution has any flaws or assumptions.

For the number to be divisible by 36, it must be divisible by 4 and 9.

An integer is divisible by 4 if its last two digits form a multiple of 4.
Here, the last two digits are 5n.
Between 50 and 59 there are two multiples of 4:
52 and 56
Thus, n=2 or n=6.

An integer is divisible by 9 if its digit sum is a multiple of 9.
Here, the digit sum = 5+m+1+5+n = 11+m+n

Case 1: n=2 --> digit sum = 11+m+2 = 13+m
The digit sum will be a multiple of 9 if m=5:
13+5 = 18
In this case, |m-n| = |5-2| = 3

Case 2: n=6 --> digit sum = 11+m+6 = 17+m
The digit sum will be a multiple of 9 if m=1:
17+1 = 18
In this case, |m-n| = |1-6| = 5

The greatest possible value for |m-n| is yielded by Case 2.

For a number to be divisible by 36, it must have two 2s and two 3s i.e. to be divisible by 4 and 9. So an even number with the last two digits divisible by 4 and the sum of digits to be divisible by 9.

The sum of digits is: 5 + m + 1 + 5 + n = 11 + m + n (divisible by 9 and the last two digits (5,n) divisible by 4)

A number is divisible by 4, if the last two digits are divisible by 4. So, that means, the number has to end with 52 or 56.

5 + m + 1 + 5 + 2 = 13 + M
5 + m + 1 + 5 + 6 = 17 + N

a number is divisible by 9 if the sum of the digits is divisible by 9. However, the question asks for | m - n | to be maximum.

5 + m + 1 + 5 + 2 = 13 + M

The max value of | m - n | in this case is possible when m-9. But the number wouldn't be divisible by 9. The maximum possible, keeping into perspective the divisibility by 9, would be when m=5. In which case | m - n | would be 3.

5 + m + 1 + 5 + 6 = 17 + N
In the other case, similarly, the maximum value of | m - n | would be when m= 0. But then, the number wouldn't be divisible by 9. So we are looking for a number that is as far from 6 as possible. 1 suffices the condition.

Hence, the maximum possible value for |m-n| = 5

Answer is C.

Shoot me if I'm incorrect. I'm not confident though. Generally, I don't get answers right.

When nothing clicks, use brute force

Divisibility rule for 36 =
Divisibility rule for 9 - sum of digits must be divisible by 9
&&
Divisibility rule for 4 - last 2 digits divisible by 4 & unit digit cannot be odd

--> it follows, m+n = 7 & n is even
\[
\begin{matrix}
n & m & divisibile? & |m -n| \\
0 & 7 &no & 7 \\
2 & 5 & yes & 3 \\
4 & 3 & no & \\
6 & 1 & yes & 5 \\
8 & -1 & NV & \\
\end{matrix}
\]


Therefore answer is C

I think answer is D and this is how I solved it!
5m51n - Since the no. is divisible by 36, it should be divisible by 2 & 3
Thus, N should be even and sum of integers should be divisible by 3 i.e m+n+5+5+1 should be multiple of 3

5+5+1 already makes 11 & thus we can look for the following multiples of 3 & keep looking for max. value of m-n
12 -> n = 0; m = 1 ; m-n = 1 (taking n=0 because we have to find maximum value of m-n)
15 -> n = 0; m = 4 ; m-n = 4
18 -> n = 0; m = 7; m-n = 7
21 -> n = 2; m = 8; m-n = 6 (n can't be zero as m has to be single digit & can't take 10 as value)

since we got max difference as 6, D is correct. please weight in my answer and suggest if went wrong anywhere.

How we can say, m will be odd?

sampad, it just happens to be true because of the numbers that are given. IMO, this question is best solved by using the divisibility rule of 4 and 9 and won't take more than a minute if you know these rules.

Did in a much simpler way (not sure if it's correct - Anyone could help here would be good)

36 = 2^2*3^2 -> max delta = 3^2-2^2 = 5

Is this a proper way of solving this, or was just a strike of luck?

Solution:

If a number is divisible by 36, then it’s divisible by 4 and 9. Recall that to be divisible by 4, the last two digits of the number have to be divisible by 4. To be divisible by 9, the sum of the digits of the number has to be divisible by 9.

Since the tens digit is 5, then n has to be either 2 or 6 since 52 and 56 are each divisible by 4. Now let’s look at these two cases.

Case 1: n = 2

If n is 2, then the sum of the digits, including n but excluding m, is 5 + 1 + 5 + 2 = 13. Therefore, we see that m must be 5 in order to the number to be divisible by 9. In this case, |m - n| = |5 - 2| = 3.

Case 2: n = 6

If n is 6, then the sum of the digits, including n but excluding m, is 5 + 1 + 5 + 6 = 17. Therefore, we see that m must be 1 in order to the number to be divisible by 9. In this case, |m - n| = |1 - 6| = 5.

Therefore, the maximum value of |m - n| is 5.

Answer: C

hey, instead of 2 and 8 for 21, we can assume n,m to be 1 and 9 and thereby making the difference 8 right?
also, if the sum is 21, how would it be divisible by 2? I cant understand why the max difference is 6?

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.